[十二省联考2019]异或粽子

[十二省联考2019]异或粽子

先吐槽一下,在考场上完全没有将这道题和超级钢琴联系起来,然后\(GG\),喜提\(60\)走人

赛后听说直接上可持久化\(Trie\)用堆维护就好了

然后自己回来又打开了超级钢琴.

这道题就有思路了

---

很明显,这道题我们可以利用前缀和优化到最大的\(k\)对数的异或和

我们想对于每一个\(i\)，我们在\(0—i-1\)去查阅它的异或最大值,用可持久化\(Trie\)可以办到,统计完答案及答案所在位置(设为\(ans_{pos}\))后将该区间分裂为\(0—ans_{pos - 1}\)和\(ans_{pos + 1}—i -1\)

剩下的以此类推，放到堆里去维护就好了

当然,因为异或第零个数比较麻烦,我将\(rt\)数组整体右移了一位。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cctype>
#include<algorithm>
#include<queue>
#define LL long long
#define pii pair<LL,int>
#define mk make_pair
using namespace std; 
const int N = 5e5 + 3;
LL a[N],sum[N];
inline LL read(){
	char ch = getchar();LL v = 0,c = 1;
	while(!isdigit(ch)){
		if(ch == '-') c = -1;
		ch = getchar();	
	}
	while(isdigit(ch)){
		v = v * 10 + ch - 48;
		ch = getchar();
	}
	return v * c;	
}
int n,k;int rt[N];
struct Trie{
	struct node{
		int lc,rc;
		int id,size;
	}a[N * 45];
	int t;
	inline void ins(int &u,int id,LL val,int dep){
		a[++t] = a[u];
		u = t;
		a[u].size++;
		if(dep == -1){a[u].id = id;return ;}
		if((1ll << dep) & val) ins(a[u].rc,id,val,dep - 1);
		else ins(a[u].lc,id,val,dep - 1);
	}
	inline pii query(int u1,int u2,LL ans,LL val,int dep){
		if(dep == -1) return mk(ans,a[u2].id);
		int lsum = a[a[u2].lc].size - a[a[u1].lc].size;
		int rsum = a[a[u2].rc].size - a[a[u1].rc].size;
		//printf("%d %d %lld %lld %d %d\n",u1,u2,ans,val,lsum,rsum);
		if((1ll << dep) & val){
			if(lsum) return query(a[u1].lc,a[u2].lc,ans | (1ll << dep),val,dep - 1);
			else return query(a[u1].rc,a[u2].rc,ans,val,dep - 1);
		}
		else{
			if(rsum) return query(a[u1].rc,a[u2].rc,ans | (1ll << dep),val,dep - 1);
			else return query(a[u1].lc,a[u2].lc,ans,val,dep - 1);
		}
	}	
}T;
struct ting{
	LL val;
	int id,from,to,ans_pos;
	ting (int idd,int fromm,int too,int ans_p,LL v){
		id = idd,from = fromm,to = too,ans_pos = ans_p;
		val = v;	
	}
};
bool operator < (const ting &x,const ting &y){
	return x.val < y.val;	
}
priority_queue <ting> q;
int main(){
	n = read(),k = read();
	for(int i = 1;i <= n;++i) a[i] = read(),sum[i] = sum[i - 1] ^ a[i];
//	for(int i = 1;i <= n;++i) printf("%lld\n",sum[i]);puts("");
	T.ins(rt[1],0,sum[0],33);
	for(int i = 1;i < n;++i){
		rt[i + 1] = rt[i];
		T.ins(rt[i + 1],i,sum[i],33);
	}
//	printf("%d\n",rt[0]);
	for(int i = 1;i <= n;++i){
		pii now = T.query(rt[0],rt[i],0,sum[i],33);
	//	printf("%lld %d\n",now.first,now.second);
		q.push(ting(i,0,i - 1,now.second,now.first));
	}
	LL ans = 0;
	while(k--){
		ting now = q.top();q.pop();
		ans += now.val;
	//	printf("%d %d %d %d %lld\n",now.id,now.from,now.to,now.ans_pos,now.val);
		if(now.ans_pos - 1 >= now.from){
			pii cnt = T.query(rt[now.from],rt[now.ans_pos],0,sum[now.id],33);
			q.push(ting(now.id,now.from,now.ans_pos - 1,cnt.second,cnt.first));	
		}
		if(now.ans_pos + 1 <= now.to){
			pii cnt = T.query(rt[now.ans_pos + 1],rt[now.to + 1],0,sum[now.id],33);
			q.push(ting(now.id,now.ans_pos + 1,now.to,cnt.second,cnt.first));	
		}
	}
	printf("%lld\n",ans);
	return 0;	
}